Question

Two bodies of equal masses are connected by a light inextensible string passing over a smooth frictionless pulley. The amount of mass that should be transferred from one to another so that both the masses mole 50 m in 5 s is

• A. 30%
• B. 40%
• C. 70%
• D. 50%

Answer 1

Answer: (B)

Since each mass is moving 50 m in 5 s, therefore using the relation $s=ut+\frac{1}{2}a{t}^2,$ we have $50=0\times 5+\frac{1}{2}\times a\times 5^2$ or $a=\frac{100}{25}=4m{s}^2$ Let mass of one become $m_1$ and that of other $m_2,$ where $m_1>m_2$ . As $m_1$ moves downwards with acceleration $a=4m{s}^{-2}$ $\therefore$ $a=\left(\frac{m_1-m_2}{m_1+m_2}\right)g$ So, $4=\left(\frac{m_1-m_2}{m_1+m_2}\right)(10)$ $\left(\frac{m_1-m_2}{m_1+m_2}\right)=\frac{4}{10}=\frac{2}{5}$ $\therefore$ % of mass transferred $=\left(\frac{m_1-m_2}{m_1+m_2}\right)\times 100$ $=\frac{2}{5}\times 100$ $=40$